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\(\frac{2018^{100}+1}{2018^{90}+1}\)= \(\frac{2018^{10}+1}{1+1}\)\(\frac{2018^{10}+1}{2}\)
\(\frac{2018^{99}+1}{2018^{89}+1}\)= \(\frac{2018^{10}+1}{1+1}\)= \(\frac{2018^{10}+1}{2}\)
=> \(\frac{2018^{100}+1}{2018^{90}+1}=\frac{2018^{99}+1}{2018^{89}+1}\)
nhớ bảo kê nha Duyên
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E =\(\frac{2018^{99}-1}{2018^{100}-1}\)so sánh với F =\(\frac{2018^{98}-1}{2018^{99}-1}\)
Ai nhanh tk
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(E=\frac{2018^{99}-1}{2018^{100}-1}\)
\(\Leftrightarrow2018E=\frac{2018^{100}-2018}{2018^{100}-1}\)
\(\Leftrightarrow2018E=1-\frac{2017}{2018^{100}-1}\) (2)
Lại có \(F=\frac{2018^{98}-1}{2018^{99}-1}\)
\(\Leftrightarrow2018F=\frac{2018^{99}-2018}{2018^{99}-1}\)
\(\Leftrightarrow2018F=1-\frac{2017}{2018^{99}-1}\) (2)
Mà \(2018^{100}>2018^{99}>0\)
\(\Leftrightarrow2018^{100}-1>2018^{99}-1\)
\(\Leftrightarrow\frac{2017}{2018^{100}-1}< \frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow-\frac{2017}{2018^{100}-1}>-\frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow1-\frac{2017}{2018-1}>1-\frac{2017}{2018^{99}-1}\) (3)
Từ (1) ;(2) và (3) <=> 2018E > 2018 F > 0
<=> E > F
Vậy E > F
@@ Học tốt
Chiyuki Fujito
K cần tk
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a ) Ta có :
\(\frac{450}{463}=1-\frac{13}{463}\) ( 1 )
\(\frac{123}{126}=1-\frac{3}{126}\)( 2 )
Từ ( 1 ) và ( 2 ) thấy 13/463 > 3/126 do đó 450/463 < 123/126
Vậy 450/463 < 123/126
b ) Ta có :
\(\frac{36}{53}=1-\frac{17}{53}\)( 1 )
\(\frac{58}{89}=1-\frac{31}{89}\)( 2 )
Từ 1 và 2 thấy 31/89 > 17/53 => 35/53 > 58/89
Vậy 35/53 > 58/89
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nhanh nha mọi người. Ngày mai mình thi rồi. Ai giải nhanh nhất mình k cho!
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\(2018^{100}+2018^{99}\)
\(=2018^{99}.\left(2018+1\right)\)
\(=2018^{99}.2019\)\(< 2019^{99}.2019=2019^{100}\)
\(\Rightarrow2018^{100}+2018^{99}< 2019^{100}\)
Vậy \(2018^{100}+2018^{99}< 2019^{100}\)
~~Hok tốt~~
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
.....................................
\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
giúp mình với